Este sábado, como les conté en mi entrada anterior, transcurrió el encuentro G4G en homenaje a Martin Gardner.
En una de las exposiciones, Rodolfo Kurchan contó como descubrió (SIN EL USO DE COMPUTADORA!!!) el cuadrado mágico de 4x4 formado por 16 números pandigitales cuya número mágico también es un número pandigital.
Suma mágica : 4120967358
Este cuadrado apareció en el número 28 de la revista "Journal of rercreational mathematics", como así también en el libro de C.A Pickover "Wonders of numbers"
La imagen la tome de la página de Carlos Rivera, PrimePuzzles
En la exposición del sábado, Rodolfo explicó que se logró un cuadrado mágico pandigital de 3x3 con una suma mágica menor, justamente en el site de Carlos Rivera y propuso encontrar el cuadrado mágico con la mayor suma mágica pandigital posible.
Yo acepté el desafio y encontré los siguientes cuadrados mágicos, ambos con suma mágica pandigital 9875304162
La pregunta es evidente ¿se puede lograr un cuadrado mágico con una suma mágica pandigital mayor?
Si lo quieres compartir o guardar
Entradas
I have found a way to get over 9,876,000,000 for the magic sum. Consider that 9,875,630,412 is relatively obvious, and so is 9,875,643,012 after that; just shuffle the digits the same way in each cell, and the magic sum is shuffled the same way.
ResponderEliminarIn each of the given magic squares, the column sums are 8, 16, 27, 4, 11, 19, 13, 10, 15, and 12. I'm assuming that the optimal solution only requires shuffling like I did above; I could be wrong, but at least I'm getting good results.
Now, including carry-overs, the column sums become 9, 18, 27, 5, 13, 20, 14, 11, 16, and 12. There might be an easy shuffle to get the magic sum over 9,876,000,000, but it involves shuffling the cell numbers differently from the sum. The obvious move is to put the sixth original sum, 19, into the fifth column, so that the carry-over of 2 turns the 4 into a 6. This creates problems, but they are manageable.
Since 4/5 in the fourth column has become 4/6, we are now missing a 5 and don't need 15/16. For that reason, I move 15 to the tenth column to make 15/15. That makes the sum 9,876,0xx,xx5 with unassigned column sums 11, 13, 10, and 12. Now it becomes clear that each x in the sum will both give and receive a carry-over of 1, regardless of order. So my result is 9,876,043,215 with column sums of 8/9, 16/18, 27, 4/6, 19/20, 13/14, 12/13, 11/12, 10/11, and 15.
Oops. My answer is wrong and breaks the diagonals.
ResponderEliminarThere's a proof that the center cell of a 3*3 magic square must be exactly 1/3 of the magic sum. That means the magic sum can't be 9,876,xxx,xxx because the center square would have to be 3,292,xxx,xxx. And here's a page that says you used the largest possible center number: http://www.iread.it/panmult.php